HSC 2015 MX2 Marathon ADVANCED (archive) (4 Viewers)

InteGrand · Sep 4, 2015

Re: HSC 2015 4U Marathon - Advanced Level

glittergal96 said:
I think you might have some quantifiers the wrong way around?

We want to find a real number a such that f(x) =< ax for all f satisfying the inequality in the original question.

In particular, we want to find the least a that works for every such f.

We know a = 1 works for the function f(x) = x, but it is perfectly conceivable that there are other functions which satisfy the conditions of the problem, but for which we don't have g(x) =< x.

$Ah right yes, I thought it meant find the lowest possible $\alpha$ that works for some $f$, and all such $f$ for which this works. My bad.$

glittergal96 · Sep 4, 2015

Re: HSC 2015 4U Marathon - Advanced Level

InteGrand said:
$Ah right yes, I thought it meant find the lowest possible $\alpha$ that works for some $f$, and all such $f$ for which this works. My bad.$

All good, looking back on the initial wording I can understand the misinterpretation.

I will edit accordingly.

glittergal96 · Sep 4, 2015

Re: HSC 2015 4U Marathon - Advanced Level

glittergal96 said:
If $f:[0,1]\rightarrow \mathbb{R}$ is a non-negative function such that

$f(1)=1$

and

$f(x+y)\geq f(x)+f(y)\quad \textrm{for }0\leq x,y\textrm{ and }x+y\leq 1$

we say that f is a sexy function.

Find the smallest real $\alpha$ such that

$f(x)\leq \alpha x$

for all $x$ and all sexy functions $f$ .

Bump with clearer wording.

dan964 · Sep 4, 2015

Re: HSC 2015 4U Marathon - Advanced Level

^ nice
that is a lot clearer.
It does also restrict the range of y between 0 and 1?
(to which the statement f(x+y)>=f(x)+f(y) holds true).

The domain of f is $0\leq x \leq 1$ (given)
The range of f is $0\leq y \leq 1$ as $x+y\leq 1$

Hence the maximum value of range between 0 and 1 is 1, which is also f(1). Therefore the curve is monotonically increasing.

glittergal96 · Sep 4, 2015

Re: HSC 2015 4U Marathon - Advanced Level

dan964 said:
It does also restrict the range of y between 0 and 1?
(to which the statement f(x+y)>=f(x)+f(y) holds true).

The statement is required to hold true for any pair of non-negative numbers x and y such that x+y =< 1.

(So their sum is still in the domain of the function.)

This condition was already included in the initial and revised wording of the question.

glittergal96 · Sep 4, 2015

Re: HSC 2015 4U Marathon - Advanced Level

dan964 said:
The range of f is $0\leq y \leq 1$ as $x+y\leq 1$

Hence the maximum value of range between 0 and 1 is 1, which is also f(1). Therefore the curve is monotonically increasing.

Uh, I think you are getting a bit confused, this quote makes very little sense.

Any sexy function f satisfies

f(x+y) >= f(x) + f(y)

for all non-negative x and y such that x+y =< 1.

x and y are just variables used for the sake of defining what it means for a function to be sexy, they don't have any relationship like y=f(x).

Monotonicity just comes from the fact that if x >= y then:

f(x) = f((x-y) + y) >= f(x-y) + f(y) >= f(y).

where the two inequalities follow from the definition of a sexy function.

dan964 · Sep 5, 2015

Re: HSC 2015 4U Marathon - Advanced Level

glittergal96 said:
Uh, I think you are getting a bit confused, this quote makes very little sense.

Any sexy function f satisfies

f(x+y) >= f(x) + f(y)

for all non-negative x and y such that x+y =< 1.

x and y are just variables used for the sake of defining what it means for a function to be sexy, they don't have any relationship like y=f(x).

Monotonicity just comes from the fact that if x >= y then:

f(x) = f((x-y) + y) >= f(x-y) + f(y) >= f(y).

where the two inequalities follow from the definition of a sexy function.

ah right, x+y<=1 doesn't necessary apply to all values of the domain as f(1)=1 is an obvious exception.

dan964 · Sep 5, 2015

Re: HSC 2015 4U Marathon - Advanced Level

glittergal96 said:
Uh, I think you are getting a bit confused, this quote makes very little sense.
Monotonicity just comes from the fact that if x >= y then:
f(x) = f((x-y) + y) >= f(x-y) + f(y) >= f(y).
where the two inequalities follow from the definition of a sexy function.

doesn't the statement x>=y imply that it must lie under the graph y=x for all values that occur with the region below (which is where the inequality holds true), and also even for values outside of this region (up to x=1)

dan964 · Sep 5, 2015

Re: HSC 2015 4U Marathon - Advanced Level

for (4) on the previous page, $y=\sqrt{x}$ is a case that does not hold
as $\sqrt{x+y} >= \sqrt{x}+\sqrt{y}$ does not hold true.
while all functions y=x^n hold, whereby n is positive hold. as (x+y)^n >= x^n + y^n which could be proven by induction.
(all of the latter for all x in the domain lie have values where y<=x)

InteGrand · Sep 5, 2015

Re: HSC 2015 4U Marathon - Advanced Level

dan964 said:
doesn't the statement x>=y imply that it must lie under the graph y=x for all values that occur with the region below (which is where the inequality holds true), and also even for values outside of this region (up to x=1)

$No, $y$ was not used as a dependent variable, it's just the name of an input to the function. If it helps, replace $x$ and $y$ with the names $x_1$ and $x_2$.$

dan964 · Sep 5, 2015

Re: HSC 2015 4U Marathon - Advanced Level

that does help, thanks
this makes sense with relation to "convexity", since it involves I guess comparing the sum of two function values.

glittergal96 · Sep 6, 2015

Re: HSC 2015 4U Marathon - Advanced Level

Here is a further hint.

If a function f satisfies f(0)=0 and is convex, then f(x+y) >= f(x) + f(y) for all x and y, as:

$f(x)+f(y) \leq \frac{x}{x+y}f(x+y)+\frac{y}{x+y}f(x+y)=f(x+y).$

dan964 · Sep 7, 2015

Re: HSC 2015 4U Marathon - Advanced Level

glittergal96 said:
Here is a further hint.

If a function f satisfies f(0)=0 and is convex, then f(x+y) >= f(x) + f(y) for all x and y, as:

$\leq \frac{x}{x+y}f(x+y)+\frac{y}{x+y}f(x+y)$ *

where did the middle section come from?
I understand where the last section came from.

f(x+h) implies directly that the value (x+h) must occur in the domain 0 to 1
The definition of convex means the secant equation must lie above the graph of f(x)
The gradient of the secant is
$\frac{f(x+h)-f(x)}{h}$

Use the point $y=0.5[f(x+h)+f(x)], x= 0.5(2x+h)$ , [otherwise the answer will f(x)]

InteGrand · Sep 7, 2015

Re: HSC 2015 4U Marathon - Advanced Level

dan964 said:
where did the middle section come from?
I understand where the last section came from.

f(x+h) implies directly that the value (x+h) must occur in the domain 0 to 1
The definition of convex means the secant equation must lie above the graph of f(x)
The gradient of the secant is
$\frac{f(x+h)-f(x)}{h}$

Use the point $y=0.5[f(x+h)+f(x)], x= 0.5(2x+h)$ , [otherwise the answer will f(x)]

$The line interval between $x_1$ and $x_2>x_1$ is the set of all $x$ such that $x=(1-t)x_1 + tx_2$ for some $t\in [0,1]$. The $y$-value of this point on the line is $(1-t)f(x_1) + f(x_2)$, if $x_1,x_2$ are in the domain of $f$. So the condition that the secant line always lies above the function is equivalent to saying $f((1-t)x_1 + tx_2)\leq (1-t)f(x_1) + tf(x_2)$ for all $t\in [0,1]$ and for all $x_1,x_2$ in the domain of $f$. The middle inequality in glittergal96's post comes from taking $t=\frac{x_1}{x_1+x_2}$ (which is clearly between $0$ and $1$ for that function)).$

(See this diagram from Wikipedia: https://upload.wikimedia.org/wikipedia/commons/c/c7/ConvexFunction.svg)

dan964 · Sep 7, 2015

Re: HSC 2015 4U Marathon - Advanced Level

thanks. sorry didn't get to finish working it out.

I suspect the answer then is alpha=1.

glittergal96 · Sep 7, 2015

Re: HSC 2015 4U Marathon - Advanced Level

Nope, alpha is not 1. Try to find an upper bound first

.

InteGrand · Sep 7, 2015

Re: HSC 2015 4U Marathon - Advanced Level

dan964 said:
so you are saying that $f(x_1)+f(x_2)=f(\frac{2f(x_1) f(x_2)}{f(x_1)+f(x_2)})$

No

dan964 · Sep 7, 2015

Re: HSC 2015 4U Marathon - Advanced Level

InteGrand said:
No

I got from using your substitution on the left hand side and comparing it with Glittergal96's inequality, that I originally asked about.

I understand how you derived it except for the relationship between the LHS and the actual inequality posed a couple of replies up.

dan964 · Sep 7, 2015

Re: HSC 2015 4U Marathon - Advanced Level

glittergal96 said:
Nope, alpha is not 1. Try to find an upper bound first .

isn't f(1)=1 and f(0)=0, so thus it is secant (y=x), the join between that, that must be above the graph if it convex?
or does convexity not apply for this case of extremities?

InteGrand · Sep 7, 2015

Re: HSC 2015 4U Marathon - Advanced Level

dan964 said:
I got from using your substitution on the left hand side and comparing it with Glittergal96's inequality, that I originally asked about.

I understand how you derived it except for the relationship between the LHS and the actual inequality posed a couple of replies up.

$That formula you posted isn't true, e.g. it doesn't work if we pick $x_1$ = 0 and $x_2=1$.$

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